Let's break down the solution step by step: --- ### **Step 1: Understanding the Problem** We are to **write a function** (in terms of \( V_t \)) for the efficiency of power transmission. As \( V_t \) increases, efficiency should approach 1. - **Given**: - Length of aluminum wire: \( L = 1\, \text{km} = 1000\, \text{m} \) - Diameter of wire: \( d = 1\, \text{cm} = 0.01\, \text{m} \) - Resistivity of aluminum: \( \rho = 2.65 \times 10^{-8}\, \Omega\cdot\text{m} \) - Load voltage: \( V_c = 110\, \text{V} \) - Total consumer power: \( P_L = 1\, \text{GW} = 1 \times 10^9\, \text{W} \) --- ### **Step 2: Find Wire Resistance \( R_w \)** \[ R_w = \rho \frac{L}{A} \] Where \( A \) is the cross-sectional area: \[ A = \pi \left(\frac{d}{2}\right)^2 = \pi (0.005)^2 = 7.85 \times 10^{-5} \text{ m}^2 \] \[ R_w = \frac{2.65 \times 10^{-8} \times 1000}{7.85 \times 10^{-5}} = 3.38 \times 10^{-4} \, \Omega \] --- ### **Step 3: Express Load Resistance \( R_L \) in terms of \( V_c \) and \( P_L \)** \[ P_L = \frac{V_c^2}{R_L} \implies R_L = \frac{V_c^2}{P_L} = \frac{(110)^2}{1 \times 10^9} = 1.21 \times 10^{-5} \, \Omega \] --- ### **Step 4: Express Efficiency** Efficiency is given by: \[ \text{efficiency} = \frac{V_c^2 / R_L}{V_g I_g} \] But \( V_g I_g \) is total power output from the generator, while \( V_c^2/R_L \) is the power consumed at the load. The total current in the system, once stepped up to transmission voltage \( V_t \) (assuming ideal transformers), is: \[ I_t = \frac{P_L}{V_t} \] The power loss in the wire is: \[ P_\text{loss} = I_t^2 R_w = \left(\frac{P_L}{V_t}\right)^2 R_w \] The total power output from the generator is: \[ P_\text{gen} = P_L + P_\text{loss} = P_L + \left(\frac{P_L}{V_t}\right)^2 R_w \] So, the efficiency is: \[ \eta = \frac{P_L}{P_L + \left(\frac{P_L}{V_t}\right)^2 R_w} \] --- ### **Step 5: Simplify the Expression** \[ \eta = \frac{1}{1 + \frac{P_L R_w}{V_t^2}} \] --- ### **Step 6: Plug in All Known Values** - \( P_L = 1 \times 10^9 \) W - \( R_w = 3.38 \times 10^{-4} \, \Omega \) \[ \eta(V_t) = \frac{1}{1 + \frac{(1 \times 10^9) \times (3.38 \times 10^{-4})}{V_t^2}} \] \[ = \frac{1}{1 + \frac{3.38 \times 10^5}{V_t^2}} \] --- ### **Step 7: Final Function** \[ \boxed{ \eta(V_t) = \frac{1}{1 + \frac{3.38 \times 10^5}{V_t^2}} } \] --- ### **Step 8: Interpretation** - As \( V_t \to \infty \), the term \( \frac{3.38 \times 10^5}{V_t^2} \to 0 \), so \( \eta \to 1 \). - **This matches the requirement:** the function shows that as \( V_t \) gets larger, efficiency approaches 1. --- ## **Summary Table** | Step | Description | Equation/Result | |------|--------------------------------------------------|--------------------------------------------------------| | 1 | Wire resistance | \( R_w = 3.38 \times 10^{-4} \, \Omega \) | | 2 | Load resistance | \( R_L = 1.21 \times 10^{-5} \, \Omega \) | | 3 | Efficiency (in terms of \( V_t \)) | \( \eta = \frac{1}{1 + \frac{3.38 \times 10^5}{V_t^2}} \) | --- **This is the function you should plot and use to show the trend as \( V_t \) increases!**